Difference between revisions of "Wooga Looga Theorem"

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=Definition=
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If there is <math>\triangle ABC</math> and points <math>D,E,F</math> on the sides <math>BC,CA,AB</math> respectively such that <math>\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r</math>, then the ratio <math>\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}</math>.
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Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
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=Proofs=
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==Proof 1==
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Proof by Gogobao:
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We have: <math>\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} </math>
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We have: <math>[DEF] = [ABC] - [DCE] - [FAE] - [FBD]</math>
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<math>[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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<math>[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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<math>[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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Therefore <math>[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})</math>
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So we have <math>\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}</math>
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==Proof 2==
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Proof by franzliszt
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Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&0&\tfrac {1}{r+1}\\  \tfrac {1}{r+1}&\tfrac {r}{r+1}&0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.</cmath>
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==Proof 3==
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Proof by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST):
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WLOG we let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math> for <math>x</math>, <math>y\in\mathbb{R}</math>. We then use Shoelace Forumla to get <math>[ABC]=\frac12|y|</math>. We then figure out that <math>D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)</math>, <math>E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)</math>, and <math>F=\left(\frac{r}{r+1}, 0\right)</math> so we know that by Shoelace Formula <math>\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|</math>. We know that <math>\frac{r^2-r+1}{(r+1)^2}\ge0</math> for all <math>r\in\mathbb{R}</math> so <math>\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}</math>.
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==Proof 4==
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Proof by ishanvannadil2008:
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Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)
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==Proof 5==
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Proof by tigerzhang:
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Consider any nondegenerate triangle <math>ABC</math>. We can apply a shear in the direction parallel to <math>\overline{BC}</math> and a distortion in the direction perpendicular to <math>\overline{BC}</math> to move <math>A</math> to any point on the plane while fixing <math>B</math> and <math>C</math>. This can define any triangle up to scaling, rotation, and orientation, so we can map points in such a way that <math>\triangle ABC</math> is equilateral. Since we have only applied linear transformations, collinear length ratios and all area ratios are constant. Thus, we can assume that <math>\triangle ABC</math> is equilateral. Also assume WLOG that its side length is <math>1</math>, so <math>AF=BD=CE=\frac{r}{r+1}</math> and <math>AE=BF=CD=\frac{1}{r+1}</math>.
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By the Law of Cosines, <cmath>EF^2=AE^2+AF^2-AE \cdot AF=\left(\frac{1}{r+1}\right)^2+\left(\frac{r}{r+1}\right)^2-\frac{1}{r+1} \cdot \frac{r}{r+1}=\frac{r^2-r+1}{(r+1)^2}.</cmath>
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Since <math>\frac{[DEF]}{[ABC]}=\left(\frac{EF}{BC}\right)^2=EF^2</math>, we have the desired result.
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=Application 1=
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==Problem==
 
The Wooga Looga Theorem states that the solution to this problem by franzliszt:
 
The Wooga Looga Theorem states that the solution to this problem by franzliszt:
  
 
In <math>\triangle ABC</math> points <math>X,Y,Z</math> are on sides <math>BC,CA,AB</math> such that <math>\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71</math>. Find the ratio of <math>[XYZ]</math> to <math>[ABC]</math>.
 
In <math>\triangle ABC</math> points <math>X,Y,Z</math> are on sides <math>BC,CA,AB</math> such that <math>\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71</math>. Find the ratio of <math>[XYZ]</math> to <math>[ABC]</math>.
  
is this solution by RedFireTruck:
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==Solution 1==
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One solution is this one by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST):
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WLOG let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math>. Then <math>[ABC]=\frac12|y|</math> by Shoelace Theorem and <math>X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)</math>, <math>Y=\left(\frac{x}{8}, \frac{y}{8}\right)</math>, <math>Z=\left(\frac78, 0\right)</math>. Then <math>[XYZ]=\frac12\left|\frac{43y}{64}\right|</math> by Shoelace Theorem. Therefore the answer is <math>\boxed{\frac{43}{64}}</math>.
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==Solution 2==
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or this solution by franzliszt:
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We apply Barycentric Coordinates w.r.t. <math>\triangle ABC</math>. Let <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then we find that <math>X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix}
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x_{1} &y_{1}  &z_{1} \\
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x_{2} &y_{2}  &z_{2} \\
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x_{3}& y_{3} & z_{3}
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\end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\frac{[XYZ]}{[ABC]}=\begin{vmatrix}
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0&\tfrac 18&\tfrac 78\\
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\tfrac 78&0&\tfrac 18\\
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\tfrac18&\tfrac78&0
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\end{vmatrix}=\frac{43}{64}.</cmath> <math>\blacksquare</math>
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==Solution 3==
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or this solution by aaja3427:
  
WLOG let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math>. Then <math>[ABC]=\frac12|y|</math> and <math>X=(\frac{7x+1}{8}, \frac{7y}{8})</math>, <math>Y=(\frac{x}{8}, \frac{y}{8})</math>, <math>Z=(\frac78, 0)</math>. Then <math>[XYZ]=\frac12|\frac{43y}{64}|</math>. Therefore the answer is <math>\boxed{\frac{43}{64}}</math>.
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According the the Wooga Looga Theorem, It is <math>\frac{49-7+1}{8^2}</math>. This is <math>\boxed{\frac{43}{64}}</math>
  
and that the solution to this problem by Matholic:
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==Solution 4==
  
The figure below shows a triangle ABC which area is 72cm2. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF
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or this solution by AoPS user ilovepizza2020:
  
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We use the <math>\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}</math> to instantly get <math>\boxed{\frac{43}{64}}</math>. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)
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==Solution 5==
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or this solution by eduD_looC:
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This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being <math>\boxed{\frac{43}{64}}</math>. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.
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==Solution 6==
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or this solution by CoolJupiter:
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Wow. All of your solutions are slow, compared to my sol:
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By math, we have <math>\boxed{\frac{43}{64}}</math>.
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~CoolJupiter
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^
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|
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EVERYONE USE THIS SOLUTION IT'S BRILLIANT
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~bsu1
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Yes, very BRILLIANT!
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~ TheAoPSLebron
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==The Best Solution==
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By the <math>1+1=\text{BREAD}</math> principle, we get <math>\boxed{\frac{43}{64}}</math>. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.
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==Easiest Solution==
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The answer is clearly <math>\boxed{\frac{43}{64}}</math>. We leave the proof and intermediate steps to the reader as an exercise.
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==Most Practical Solution==
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Refer to Sun Tzu's "Art of War", page <math>93865081</math>: "The answer to the problem is <math>\boxed{\frac{43}{64}}</math>". Discovered by mutinykids, who read the entire book. If you pay him $5 he will give more wise advice.
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=Application 2=
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==Problem==
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The Wooga Looga Theorem states that the solution to this problem by Matholic:
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The figure below shows a triangle ABC whose area is <math>72 \text{cm}^2</math>. If <math>\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}</math>, find <math>[DEF].</math>
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~LaTeX-ifyed by RP3.1415
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==Solution 1==
 
is this solution by franzliszt:
 
is this solution by franzliszt:
  
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\tfrac16&0&\tfrac56
 
\tfrac16&0&\tfrac56
 
\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math>
 
\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math>
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==Solution 2==
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or this solution by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST):
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By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}</math>. We are given that <math>[ABC]=72</math> so <math>[DEF]=\frac{7}{12}\cdot72=\boxed{42}</math>
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=Application 3=
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==Problem==
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The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:
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Find the ratio <math>\frac{[GHI]}{[ABC]}</math> if <math>\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12</math> and <math>\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1</math> in the diagram below.<asy>
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draw((0, 0)--(6, 0)--(4, 3)--cycle);
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draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);
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draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);
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label("$A$", (0, 0), SW);
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label("$B$", (6, 0), SE);
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label("$C$", (4, 3), N);
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label("$D$", (2, 0), S);
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label("$E$", (16/3, 1), NE);
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label("$F$", (8/3, 2), NW);
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label("$G$", (11/3, 1/2), SE);
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label("$H$", (4, 3/2), NE);
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label("$I$", (7/3, 1), W);
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</asy>
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==Solution 1==
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is this solution by franzliszt:
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By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13</math>. Notice that <math>\triangle GHI</math> is the medial triangle of '''Wooga Looga Triangle ''' of <math>\triangle ABC</math>. So <math>\frac{[GHI]}{[DEF]}=\frac 14</math> and <math>\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}</math> by Chain Rule ideas.
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==Solution 2==
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or this solution by franzliszt:
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Apply Barycentrics w.r.t. <math>\triangle ABC</math> so that <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then <math>D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)</math>. And <math>G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)</math>.
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In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\  x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.</cmath>
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=Application 4=
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==Problem==
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Let <math>ABC</math> be a triangle and <math>D,E,F</math> be points on sides <math>BC,AC,</math> and <math>AB</math> respectively. We have that <math>\frac{BD}{DC} = 3</math> and similar for the other sides. If the area of triangle <math>ABC</math> is <math>16</math>, then what is the area of triangle <math>DEF</math>? (By ilovepizza2020)
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==Solution 1==
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By Franzliszt
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By Wooga Looga, <math>\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}</math> so the answer is <math>\boxed7</math>.
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==Solution 2==
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By franzliszt
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Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\  x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&\tfrac 14&\tfrac 34\\ \tfrac 34&0&\tfrac 14\\ \tfrac 14&\tfrac 34&0 \end{vmatrix}=\frac{7}{16}.</cmath>So the answer is <math>\boxed{7}</math>.
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==Solution 3==
  
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A long story short, the answer must be <math>\boxed{7}</math> by the inverse of the Inverse Wooga Looga Theorem
  
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=Testimonials=
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Pogpr0 = wooga looga - Ladka13
 
The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.
 
The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.
 
~ilp2020
 
~ilp2020
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Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
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Franzlist is wooga looga howsopro - volkie boy
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this is in fact a pretty sensible theorem. Nothing to be so excited about, though.  ~DofL
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The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111  -centslordm
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The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 11:00, 1 February 2021 (EST)
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The Wooga Looga Theorem is the best. -aaja3427
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The Wooga Looga Theorem is needed for everything and it is great-hi..
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The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT
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This theorem has helped me with school and I am no longer failing my math class. -mchang
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I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur
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"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter
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Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)
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Too powerful... ~franzliszt
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The Wooga Looga Theorem is so pro ~ ac142931
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It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)
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This theorem changed my life... ~ samrocksnature
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Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3
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It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3
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This is franzliszt and I endorse this theorem. ~franzliszt
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This theorem is too OP. ~bestzack66
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This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283
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Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415
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The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020
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It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321
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The Wooga Looga Theorem is a special case of [https://en.wikipedia.org/wiki/Routh%27s_theorem Routh's Theorem.] So this wiki article is DEFINITELY needed. -peace
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I actually thought this was a joke theorem until I read this page - HumanCalculator9
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I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825
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This is <i>almost</i> as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun
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<s>I ReAlLy don't get it - Senguamar</s> HOW DARE YOU!!!!
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The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.
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thos theroem is very prO ~ themathboi101
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You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan
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If only I knew this on some contests that I had done previously... - JacobJB
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The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri
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"The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 21:56, 21 December 2020 (EST)
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The Wooga Looga Theorem is so OP BRUH
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thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie
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I have no idea what is going on here - awesomeguy856
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fuzimiao2013 waz hear
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this theorem is bad
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poggers theorem - awesomeming327
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The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314
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person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun
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Wooga Looga Theorem is TRASH.
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HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY  PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USA(J)MO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.
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Now I know how @Louis_Vuitton got so much smarter than me! :rotfl:
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this is the shoddiest theorem i have ever seen
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orz theorem - tigerzhang
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This theorem is very useful and good, despite its slightly meme qualities. It can be used to get high scores on any mathematics competition, proof or otherwise. - dineshs
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The number one theorem in the world, don't deny it. the best theorem THE BEST. you will get 7 points on the IMO by using this - kante314
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I got a gold medal on the 2022 winter olympics because of this theorem ~bronzetruck2016, 2021

Latest revision as of 22:58, 1 August 2021

Definition

If there is $\triangle ABC$ and points $D,E,F$ on the sides $BC,CA,AB$ respectively such that $\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r$, then the ratio $\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}$.


Created by Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

Proofs

Proof 1

Proof by Gogobao:

We have: $\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1}$

We have: $[DEF] = [ABC] - [DCE] - [FAE] - [FBD]$

$[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}$

Therefore $[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})$

So we have $\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}$

Proof 2

Proof by franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that \[\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&0&\tfrac {1}{r+1}\\   \tfrac {1}{r+1}&\tfrac {r}{r+1}&0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.\]

Proof 3

Proof by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG we let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$ for $x$, $y\in\mathbb{R}$. We then use Shoelace Forumla to get $[ABC]=\frac12|y|$. We then figure out that $D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)$, $E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)$, and $F=\left(\frac{r}{r+1}, 0\right)$ so we know that by Shoelace Formula $\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|$. We know that $\frac{r^2-r+1}{(r+1)^2}\ge0$ for all $r\in\mathbb{R}$ so $\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}$.

Proof 4

Proof by ishanvannadil2008:

Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)

Proof 5

Proof by tigerzhang:

Consider any nondegenerate triangle $ABC$. We can apply a shear in the direction parallel to $\overline{BC}$ and a distortion in the direction perpendicular to $\overline{BC}$ to move $A$ to any point on the plane while fixing $B$ and $C$. This can define any triangle up to scaling, rotation, and orientation, so we can map points in such a way that $\triangle ABC$ is equilateral. Since we have only applied linear transformations, collinear length ratios and all area ratios are constant. Thus, we can assume that $\triangle ABC$ is equilateral. Also assume WLOG that its side length is $1$, so $AF=BD=CE=\frac{r}{r+1}$ and $AE=BF=CD=\frac{1}{r+1}$.

By the Law of Cosines, \[EF^2=AE^2+AF^2-AE \cdot AF=\left(\frac{1}{r+1}\right)^2+\left(\frac{r}{r+1}\right)^2-\frac{1}{r+1} \cdot \frac{r}{r+1}=\frac{r^2-r+1}{(r+1)^2}.\] Since $\frac{[DEF]}{[ABC]}=\left(\frac{EF}{BC}\right)^2=EF^2$, we have the desired result.

Application 1

Problem

The Wooga Looga Theorem states that the solution to this problem by franzliszt:

In $\triangle ABC$ points $X,Y,Z$ are on sides $BC,CA,AB$ such that $\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71$. Find the ratio of $[XYZ]$ to $[ABC]$.

Solution 1

One solution is this one by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$. Then $[ABC]=\frac12|y|$ by Shoelace Theorem and $X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)$, $Y=\left(\frac{x}{8}, \frac{y}{8}\right)$, $Z=\left(\frac78, 0\right)$. Then $[XYZ]=\frac12\left|\frac{43y}{64}\right|$ by Shoelace Theorem. Therefore the answer is $\boxed{\frac{43}{64}}$.

Solution 2

or this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that \[\frac{[XYZ]}{[ABC]}=\begin{vmatrix} 0&\tfrac 18&\tfrac 78\\ \tfrac 78&0&\tfrac 18\\ \tfrac18&\tfrac78&0 \end{vmatrix}=\frac{43}{64}.\] $\blacksquare$

Solution 3

or this solution by aaja3427:

According the the Wooga Looga Theorem, It is $\frac{49-7+1}{8^2}$. This is $\boxed{\frac{43}{64}}$

Solution 4

or this solution by AoPS user ilovepizza2020:

We use the $\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}$ to instantly get $\boxed{\frac{43}{64}}$. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)

Solution 5

or this solution by eduD_looC:

This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being $\boxed{\frac{43}{64}}$. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.

Solution 6

or this solution by CoolJupiter:

Wow. All of your solutions are slow, compared to my sol:

By math, we have $\boxed{\frac{43}{64}}$.

~CoolJupiter ^ | EVERYONE USE THIS SOLUTION IT'S BRILLIANT ~bsu1

Yes, very BRILLIANT!

~ TheAoPSLebron

The Best Solution

By the $1+1=\text{BREAD}$ principle, we get $\boxed{\frac{43}{64}}$. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.

Easiest Solution

The answer is clearly $\boxed{\frac{43}{64}}$. We leave the proof and intermediate steps to the reader as an exercise.

Most Practical Solution

Refer to Sun Tzu's "Art of War", page $93865081$: "The answer to the problem is $\boxed{\frac{43}{64}}$". Discovered by mutinykids, who read the entire book. If you pay him $5 he will give more wise advice.

Application 2

Problem

The Wooga Looga Theorem states that the solution to this problem by Matholic:

The figure below shows a triangle ABC whose area is $72 \text{cm}^2$. If $\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}$, find $[DEF].$

~LaTeX-ifyed by RP3.1415

Solution 1

is this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\  x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that\[\frac{[DEF]}{[72]}=\begin{vmatrix} \tfrac 56&\tfrac 16&0\\ 0&\tfrac 56&\tfrac 16\\ \tfrac16&0&\tfrac56 \end{vmatrix}=\frac{7}{12}\] so $[DEF]=42$. $\blacksquare$

Solution 2

or this solution by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}$. We are given that $[ABC]=72$ so $[DEF]=\frac{7}{12}\cdot72=\boxed{42}$

Application 3

Problem

The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:

Find the ratio $\frac{[GHI]}{[ABC]}$ if $\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12$ and $\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1$ in the diagram below.[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label("$A$", (0, 0), SW); label("$B$", (6, 0), SE); label("$C$", (4, 3), N); label("$D$", (2, 0), S); label("$E$", (16/3, 1), NE); label("$F$", (8/3, 2), NW); label("$G$", (11/3, 1/2), SE); label("$H$", (4, 3/2), NE); label("$I$", (7/3, 1), W); [/asy]

Solution 1

is this solution by franzliszt:

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13$. Notice that $\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\triangle ABC$. So $\frac{[GHI]}{[DEF]}=\frac 14$ and $\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}$ by Chain Rule ideas.

Solution 2

or this solution by franzliszt:

Apply Barycentrics w.r.t. $\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)$. And $G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)$.

In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that\[\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.\]

Application 4

Problem

Let $ABC$ be a triangle and $D,E,F$ be points on sides $BC,AC,$ and $AB$ respectively. We have that $\frac{BD}{DC} = 3$ and similar for the other sides. If the area of triangle $ABC$ is $16$, then what is the area of triangle $DEF$? (By ilovepizza2020)

Solution 1

By Franzliszt

By Wooga Looga, $\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}$ so the answer is $\boxed7$.

Solution 2

By franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that\[\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&\tfrac 14&\tfrac 34\\ \tfrac 34&0&\tfrac 14\\ \tfrac 14&\tfrac 34&0 \end{vmatrix}=\frac{7}{16}.\]So the answer is $\boxed{7}$.

Solution 3

A long story short, the answer must be $\boxed{7}$ by the inverse of the Inverse Wooga Looga Theorem

Testimonials

Pogpr0 = wooga looga - Ladka13 The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020

Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

Franzlist is wooga looga howsopro - volkie boy

this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL

The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm

The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck (talk) 11:00, 1 February 2021 (EST)

The Wooga Looga Theorem is the best. -aaja3427

The Wooga Looga Theorem is needed for everything and it is great-hi..

The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT

This theorem has helped me with school and I am no longer failing my math class. -mchang

I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur

"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter

Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)

Too powerful... ~franzliszt

The Wooga Looga Theorem is so pro ~ ac142931

It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)

This theorem changed my life... ~ samrocksnature

Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3

It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3

This is franzliszt and I endorse this theorem. ~franzliszt

This theorem is too OP. ~bestzack66

This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283

Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415

The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020

It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321

The Wooga Looga Theorem is a special case of Routh's Theorem. So this wiki article is DEFINITELY needed. -peace

I actually thought this was a joke theorem until I read this page - HumanCalculator9

I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825

This is almost as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun

I ReAlLy don't get it - Senguamar HOW DARE YOU!!!!

The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.

thos theroem is very prO ~ themathboi101

You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan

If only I knew this on some contests that I had done previously... - JacobJB

The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri

"The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ Aops-g5-gethsemanea2 (talk) 21:56, 21 December 2020 (EST)

The Wooga Looga Theorem is so OP BRUH

thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie

I have no idea what is going on here - awesomeguy856

fuzimiao2013 waz hear

this theorem is bad

poggers theorem - awesomeming327

The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314

person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun

Wooga Looga Theorem is TRASH.

HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USA(J)MO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.

Now I know how @Louis_Vuitton got so much smarter than me! :rotfl:

this is the shoddiest theorem i have ever seen

orz theorem - tigerzhang

This theorem is very useful and good, despite its slightly meme qualities. It can be used to get high scores on any mathematics competition, proof or otherwise. - dineshs

The number one theorem in the world, don't deny it. the best theorem THE BEST. you will get 7 points on the IMO by using this - kante314

I got a gold medal on the 2022 winter olympics because of this theorem ~bronzetruck2016, 2021

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